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AdvancedGrades 10–1125 min44 / 82

Exponential functions, equations and inequalities

The exponential function y = aˣ, its graph and properties, the number e, growth and decay models (compound interest, doubling, half-life), four methods for exponential equations, and exponential inequalities.

Check yourself
In this lesson you will learn
  • Sketch y = aˣ, state its properties and compare powers
  • Solve compound interest, doubling and half-life problems with an exponential model
  • Solve exponential equations by a common base, factoring out, substitution and the homogeneous-equation method
  • Solve exponential inequalities, taking into account whether the base is greater or less than 1

Aysel puts 1000 manat in a bank that adds 10% to the balance every year. In the first year she earns 100 manat, in the second already 110 manat, because the interest is charged on the new balance. Each year the balance is multiplied by the same number, 1.1, so after n years it is 1000 · 1.1ⁿ manat. After 10 years that is 2593.74 manat, while a plain “+100 manat a year” would give only 2000. Dependences in which the variable sits in the exponent are described by the exponential function.

In the lessons “Powers with integer exponents” and “Square roots, nth roots and rational exponents” we studied aⁿ for integer and rational exponents. Now the variable moves into the exponent: first we study the properties of this function, then the methods for solving exponential equations and inequalities. The next lesson, “Logarithms, their properties and the logarithmic function”, introduces a new operation that solves equations such as 2ˣ = 5.

Interactive
Loading simulation…
Each year the balance is multiplied by the same number (1.1): this is exponential growth. Change the interest rate and see how many years it takes the balance to double.

The exponential function and its graph

Definition
Exponential function

A function of the form y = aˣ, where the base a is a constant (a > 0, a ≠ 1) and the variable x is in the exponent. Examples: y = 2ˣ, y = (1/3)ˣ, y = 10ˣ, y = eˣ. By contrast, y = x² is a power function, not an exponential one: there the variable is in the base.

These conditions are not arbitrary. With a < 0, values such as (−4)^(1/2) = √(−4) would make no sense, and with a = 1 we only get the constant 1ˣ = 1. For a > 0, aˣ makes sense for every real x: rational exponents are defined through roots, irrational ones by approximation — 2^√2 ≈ 2.665 is the number that the powers 2^1.4, 2^1.41, 2^1.414, … approach.

y = aˣ (a > 0, a ≠ 1)
where:
  • athe base, a fixed positive number other than 1
  • xthe exponent, any real number
  • ythe value of the function, always positive

The exponential function: domain — all real numbers, range — (0, +∞). For a > 1 it is increasing, for 0 < a < 1 decreasing.

xy = 2ˣy = (1/2)ˣ
−31/88
−21/44
−11/22
011
121/2
241/4
381/8
2ˣ grows and (1/2)ˣ shrinks; both equal 1 at x = 0 and are never 0 or negative. Since (1/2)ˣ = 2⁻ˣ, the two columns are mirror images of each other.
  • Domain: all real numbers; range: (0, +∞), that is, aˣ > 0. So 2ˣ = −4 and 3ˣ = 0 have no solutions.
  • The graph passes through (0, 1) and (1, a), because a⁰ = 1 and a¹ = a.
  • For a > 1 the function is increasing, for 0 < a < 1 it is decreasing.
  • The x-axis (y = 0) is a horizontal asymptote: the graph comes as close to it as we like but never touches it.
  • The graphs of y = aˣ and y = (1/a)ˣ are symmetric about the y-axis, because (1/a)ˣ = a⁻ˣ.
Interactive
Loading simulation…
The first graph is y = aˣ, the second y = (1/a)ˣ. Move a: both graphs pass through (0, 1), never go below the x-axis and are mirror images in the y-axis. At a = 1 both turn into the line y = 1.

Monotonicity lets us compare powers without computing them: for a > 1 the larger exponent gives the larger power, for 0 < a < 1 it is the other way round. If the bases differ, first rewrite them with the same base.

Comparing powers and finding a range

Compare:
1) 2^0.7 and 2^0.5
2) 0.3⁻² and 0.3⁻¹
3) 4^0.3 and 8^0.25
4) (1/3)^√2 and (1/3)^1.5
5) Find the range of y = 2ˣ + 3.

Show solution
1) The base 2 > 1, the function is increasing; 0.7 > 0.5 ⇒ 2^0.7 > 2^0.5.
2) The base 0.3 < 1, the function is decreasing; −2 < −1 ⇒ 0.3⁻² > 0.3⁻¹ (check: 11.1 > 3.3).
3) Use the same base: 4^0.3 = 2^0.6, 8^0.25 = 2^0.75. 0.6 < 0.75 ⇒ 4^0.3 < 8^0.25.
4) √2 ≈ 1.414 < 1.5 and the base 1/3 < 1 ⇒ (1/3)^√2 > (1/3)^1.5.
5) 2ˣ takes every positive value, so 2ˣ + 3 > 3. Range: (3, +∞).

Growth and decay models. The number e

Aysel’s deposit follows this law: over equal time intervals the quantity is multiplied by the same number. If the multiplier is greater than 1, this is exponential growth (bacteria multiplying, rising prices); if it is less than 1, it is exponential decay (a car losing value, radioactive decay). Unlike linear growth, where the same amount is added each time, here the quantity grows by the same factor.

N(t) = N₀ · a^(t/T)
where:
  • N₀the initial amount (at t = 0)
  • athe multiplier per period: a = 1 + p/100 for growth by p%, a = 1 − p/100 for a decrease by p%, a = 2 for doubling, a = 1/2 for halving
  • Tthe length of one period (year, hour, day…)
  • tthe time elapsed; t/T is the number of periods

Every time T passes, the quantity is multiplied by a. Compound interest: S = S₀ · (1 + p/100)ⁿ (T = 1 year, n = number of years). Radioactive decay: m = m₀ · (1/2)^(t/T), where T is the half-life.

Interest, depreciation, multiplication and decay

1) Aysel deposits 1000 manat at 10% compound interest per year for 3 years. How much will she have at the end?
2) A car worth 20 000 manat loses 20% of its value every year. What is it worth after 3 years?
3) Bacteria double every 20 minutes. How many bacteria come from 500 bacteria after 2 hours?
4) The half-life of iodine-131 is about 8 days. How much of 80 mg of iodine is left after 24 days?

Show solution
1) a = 1.1, n = 3: S = 1000 · 1.1³ = 1000 · 1.331 = 1331 manat.
2) a = 1 − 0.2 = 0.8: 20 000 · 0.8³ = 20 000 · 0.512 = 10 240 manat.
3) T = 20 min, t = 120 min ⇒ t/T = 6: N = 500 · 2⁶ = 500 · 64 = 32 000.
4) t/T = 24 ÷ 8 = 3: m = 80 · (1/2)³ = 80 ÷ 8 = 10 mg.

What happens if interest is added more often than once a year? Suppose 1 manat is deposited at 100% a year and the interest is added n times a year, each time at the rate 100%/n. At the end of the year there are (1 + 1/n)ⁿ manat. As n grows, this number grows too, but not without bound: it approaches a certain number.

Interest addedn(1 + 1/n)ⁿ
once a year12
every half year22.25
monthly122.613035…
daily3652.714567…
a million times1 000 0002.718280…
The gain grows, but more and more slowly: the numbers approach e ≈ 2.71828.
(1 + 1/n)ⁿ → e ≈ 2.71828 (n → ∞)(1 + 1/n)ⁿ → e ≈ 2.71828 (n → ∞)
where:
  • eEuler’s number, an irrational number and the base of the natural logarithm
  • nhow many times a year the interest is added

The function y = eˣ with base e is often called simply the exponential function. When a quantity changes at every moment at a rate proportional to its size (bacteria, populations, radioactive matter), it is expressed through powers of e.

N(t) = N₀ · e^(kt)
where:
  • N₀the initial amount
  • kthe rate constant: k > 0 for growth, k < 0 for decay
  • ttime

The model of continuous growth and decay. Powers of e are found with a calculator: e ≈ 2.718, e⁻¹ ≈ 0.368, e^0.1 ≈ 1.105.

Models with base e

1) A lab has 2000 bacteria and N(t) = 2000 · e^(0.5t) (t in hours). How many bacteria are there after 2 hours?
2) The mass of a substance decreases as m(t) = 100 · e^(−0.1t) grams (t in days). How much is left after 10 days?
3) 1000 manat is kept for 1 year at 10% continuous interest: S = 1000 · e^0.1. Compare with interest added once a year.

Show solution
1) N(2) = 2000 · e^(0.5 · 2) = 2000 · e ≈ 2000 · 2.718 ≈ 5437.
2) m(10) = 100 · e⁻¹ ≈ 100 · 0.368 ≈ 36.8 g.
3) S = 1000 · e^0.1 ≈ 1105.17 manat, while once a year gives 1000 · 1.1 = 1100 manat. Continuous interest adds only 5.17 manat more: the growth is not unlimited, e caps it.

Exponential equations: four main methods

Definition
Exponential equation

An equation with the unknown in the exponent: 2^(x + 1) = 32, 9ˣ − 4 · 3ˣ + 3 = 0. Every method has the same goal: to bring the equation to the form a^f(x) = a^g(x).

a^f(x) = a^g(x) ⇔ f(x) = g(x) (a > 0, a ≠ 1)
where:
  • athe same base on both sides
  • f(x), g(x)the expressions in the exponents

Why it holds: aˣ is monotonic and takes each value only once, so equal powers have equal exponents. The equation aˣ = b has exactly one solution for b > 0 and none for b ≤ 0.

  1. 1
    Look at the bases

    Try to write all powers with one base: 4 = 2², 0.25 = 2⁻², √2 = 2^(1/2), 1/27 = 3⁻³.

  2. 2
    Several terms with one base

    Factor out the power with the smallest exponent: 3^(x + 2) − 3ˣ = 3ˣ(9 − 1).

  3. 3
    a^(2x) together with aˣ

    Substitute t = aˣ, write down t > 0 and solve the quadratic equation.

  4. 4
    Two different bases: 4ˣ, 6ˣ, 9ˣ

    It is a homogeneous equation: divide by 9ˣ and put t = (2/3)ˣ.

  5. 5
    Check

    Substitute the root into the original equation, or at least check that t > 0.

Method 1: a common base

Solve:
1) 2^(x + 1) = 32
2) 9ˣ = 27
3) (1/5)ˣ = 125
4) 4^(x² − x) = 16
5) 2ˣ = 4√2

Show solution
1) 32 = 2⁵ ⇒ x + 1 = 5 ⇒ x = 4.
2) 9ˣ = 3^(2x), 27 = 3³ ⇒ 2x = 3 ⇒ x = 3/2.
3) 1/5 = 5⁻¹ ⇒ 5^(−x) = 5³ ⇒ x = −3.
4) 16 = 4² ⇒ x² − x = 2 ⇒ x² − x − 2 = 0 ⇒ x = 2 or x = −1.
5) 4√2 = 2² · 2^(1/2) = 2^(5/2) ⇒ x = 5/2.
a^(x + k) = aˣ · aᵏ
where:
  • ka constant (the difference of the exponents)

Powers whose exponents differ by a constant share the factor aˣ. Factor it out and an ordinary number is left in the brackets.

Method 2: factoring out

Solve:
1) 3^(x + 2) − 3ˣ = 72
2) 2^(x + 1) + 2^(x − 1) = 20
3) 5ˣ + 5^(x + 1) = 150

Show solution
1) 3^(x + 2) = 9 · 3ˣ ⇒ 3ˣ(9 − 1) = 72 ⇒ 3ˣ = 9 ⇒ x = 2.
2) The smallest exponent is x − 1: 2^(x − 1)(2² + 1) = 20 ⇒ 2^(x − 1) = 4 ⇒ x − 1 = 2 ⇒ x = 3.
Check: 2⁴ + 2² = 16 + 4 = 20.
3) 5ˣ(1 + 5) = 150 ⇒ 5ˣ = 25 ⇒ x = 2.
a^(2x) = (aˣ)² = t², t = aˣ > 0
where:
  • tthe new variable; always positive, because aˣ > 0

An equation made of a^(2x), aˣ and a constant becomes a quadratic equation in t. Since a^(−x) = 1/t, equations with aˣ + a^(−x) are solved the same way.

Method 3: substitution

Solve:
1) 9ˣ − 4 · 3ˣ + 3 = 0
2) 4ˣ − 2ˣ − 12 = 0
3) 2ˣ + 2^(2 − x) = 5

Show solution
1) 9ˣ = (3ˣ)². t = 3ˣ > 0: t² − 4t + 3 = 0 ⇒ t = 1 or t = 3.
3ˣ = 1 ⇒ x = 0; 3ˣ = 3 ⇒ x = 1. Answer: 0 and 1.
2) t = 2ˣ > 0: t² − t − 12 = 0 ⇒ t = 4 or t = −3.
Reject t = −3, because 2ˣ cannot be negative. 2ˣ = 4 ⇒ x = 2.
3) 2^(2 − x) = 4/2ˣ. t = 2ˣ > 0: t + 4/t = 5 ⇒ t² − 5t + 4 = 0 ⇒ t = 1 or t = 4.
Answer: x = 0 and x = 2.
A · a^(2x) + B · aˣbˣ + C · b^(2x) = 0 | : b^(2x) ⇒ A · t² + B · t + C = 0, t = (a/b)ˣ > 0A · a^(2x) + B · aˣbˣ + C · b^(2x) = 0 | : b^(2x) ⇒ A · t² + B · t + C = 0, t = (a/b)ˣ > 0
where:
  • a, btwo different positive bases (for example 2 and 3)
  • A, B, Cnumerical coefficients

A homogeneous equation: every term has the same “degree” (a^(2x), aˣbˣ, b^(2x)). Since b^(2x) is never 0, we may divide by it and get a quadratic equation in one unknown. The simplest case: aˣ = bˣ ⇔ (a/b)ˣ = 1 ⇔ x = 0.

Method 4: homogeneous equations

Solve:
1) 5ˣ = 7ˣ
2) 3 · 4ˣ − 5 · 6ˣ + 2 · 9ˣ = 0
3) 9ˣ + 6ˣ = 2 · 4ˣ

Show solution
1) Divide by 7ˣ > 0: (5/7)ˣ = 1 = (5/7)⁰ ⇒ x = 0.
2) 4ˣ = (2ˣ)², 9ˣ = (3ˣ)², 6ˣ = 2ˣ · 3ˣ. Divide by 9ˣ: 3 · (2/3)^(2x) − 5 · (2/3)ˣ + 2 = 0.
t = (2/3)ˣ > 0: 3t² − 5t + 2 = 0 ⇒ t = 1 or t = 2/3.
(2/3)ˣ = 1 ⇒ x = 0; (2/3)ˣ = 2/3 ⇒ x = 1. Answer: 0 and 1.
Check (x = 1): 12 − 30 + 18 = 0.
3) Divide by 4ˣ: (3/2)^(2x) + (3/2)ˣ − 2 = 0. t = (3/2)ˣ > 0: t² + t − 2 = 0 ⇒ t = 1 or t = −2 (rejected).
(3/2)ˣ = 1 ⇒ x = 0.

Exponential inequalities

In an inequality we also bring both sides to one base, but when we move to the exponents one question decides everything: is the function increasing or decreasing? An increasing function gives a larger value for a larger argument, so the sign stays; for a decreasing function the sign flips.

a > 1: a^f(x) > a^g(x) ⇔ f(x) > g(x); 0 < a < 1: a^f(x) > a^g(x) ⇔ f(x) < g(x)
where:
  • athe common base
  • f(x), g(x)the expressions in the exponents

If the base is less than 1, the sign flips. If the right-hand side is negative or zero, nothing needs solving: aˣ > −3 holds for every x, and aˣ < 0 never holds.

Inequalities with a common base

Solve:
1) 2^(x − 1) > 8
2) (1/3)ˣ ≥ 27
3) 0.5^(x² − 2x) ≥ 1/8
4) 3^(x²) < 9^(x + 4)

Show solution
1) 8 = 2³, the base 2 > 1, the sign stays: x − 1 > 3 ⇒ x > 4.
2) 27 = (1/3)⁻³, the base 1/3 < 1, the sign flips: x ≤ −3.
3) 1/8 = 0.5³, the base 0.5 < 1: x² − 2x ≤ 3 ⇒ (x − 3)(x + 1) ≤ 0 ⇒ x ∈ [−1, 3].
4) 9^(x + 4) = 3^(2x + 8), the base 3 > 1: x² < 2x + 8 ⇒ (x − 4)(x + 2) < 0 ⇒ x ∈ (−2, 4).
The last two were finished with the interval method from the lesson “Quadratic inequalities and the interval method”.
Inequalities by substitution

Solve:
1) 4ˣ − 5 · 2ˣ + 4 < 0
2) 9ˣ − 2 · 3ˣ − 3 > 0

Show solution
1) t = 2ˣ > 0: t² − 5t + 4 < 0 ⇒ (t − 1)(t − 4) < 0 ⇒ 1 < t < 4.
2⁰ < 2ˣ < 2² and the base 2 > 1 ⇒ 0 < x < 2.
2) t = 3ˣ > 0: (t − 3)(t + 1) > 0. Since t > 0, t + 1 > 0, so t − 3 > 0, that is, t > 3.
3ˣ > 3¹ ⇒ x > 1.

Key points

  • y = aˣ (a > 0, a ≠ 1): domain — all real numbers, range — (0, +∞); the graph passes through (0, 1).
  • For a > 1, aˣ is increasing, for 0 < a < 1 decreasing; the graphs of y = aˣ and y = (1/a)ˣ are symmetric about the y-axis.
  • Growth and decay model: N(t) = N₀ · a^(t/T); e ≈ 2.718 is the number that (1 + 1/n)ⁿ approaches; the continuous model is N₀ · e^(kt).
  • a^f(x) = a^g(x) ⇔ f(x) = g(x). Methods: a common base, factoring out, the substitution t = aˣ > 0, dividing a homogeneous equation by b^(2x).
  • In an inequality the sign stays for a > 1 and flips for 0 < a < 1; remember that aˣ > 0.

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
What is the range of y = aˣ (a > 0, a ≠ 1)?